
#1
May1410, 09:59 AM

Sci Advisor
P: 869

Hi,
According to eg Nakahara's conventions the Laplacian on a form K is given by [tex] \Delta K = (dd^{\dagger} + d^{\dagger}d)K [/tex] In my case K is a two form living in R^3. I've calculated the Laplacian and arrive at [tex] \Delta K = \Bigl( \frac{1}{3!}\epsilon^{klm}\epsilon^n_{\ ij}\partial_k \partial_n K_{lm}  \frac{1}{4}\partial_{i}\partial^k K_{jk} \Bigr) dx^i \wedge dx^j [/tex] However, the answer seems a little odd to me. Does anyone have a reference where Laplacians of two forms are evaluated, or some comment? Thanks in forward! :) 



#2
May1410, 02:36 PM

Sci Advisor
P: 869

And another question: is on a compact manifold a harmonic form neccesarily constant, or is this only true for 0forms (functions)?




#3
May1710, 03:24 PM

P: 491

How would you go about defining a constant form?
The obvious answer to this would be that you can't define it in a coordinateinvariant way. The generalization of the idea of constant functions to forms is with the definition: a constant function is d and d* closed. 



#4
May2710, 08:58 AM

P: 3

Laplacian of a 2form
For your first question, if the codifferential is defined with a sign (1)^n(p+1), it can be shown that in euclidean space in general laplace beltrami of a form is the same as the usual old laplacian of every function factor (times the wedged covectors). It is a lengthy calculation and took me some hours, but it works ;)




#5
May2810, 03:36 PM

Sci Advisor
P: 869

Hi,
but that would mean that my calculation isn't right? About this "constant form": that should be zero, ofcourse; if a tensor is zero in one coordinate system, it's zero in every coordinate system. My bad ;) 


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